Question

Above the triangle with vertices (0,2),(1,1) and (2,2) , and under the surface z = xy.

Short answer

The required solution is $\frac{5}{3}$

Step-by-step solution

Definition of double integral

The double integral of f(x,y)over the area Ain the (x,y)plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dxdy}$

Calculation of the volume under the curve

The volume above the triangle with vertices (0,2), (1,1) and (2,2), and under the surface z = xy .

First, integrate over z :

$$\begin{gathered} |={\int }_{\mathrm{x}=1}^2{\int }_{\mathrm{y}=2-\mathrm{x}}^{\mathrm{x}}\left[{\int }_{\mathrm{z}=0}^{\mathrm{xy}}\mathrm{dz}\right]\mathrm{dxdy} \\ ={\int }_1^2\mathrm{dx}{\int }_{2-\mathrm{x}}^{\mathrm{x}}\mathrm{dyxy} \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over y:

$$\begin{gathered} |={\int }_1^2\mathrm{xdx}{\overline{)\left(\frac{1}{2}{\mathrm{y}}^2\right)}}_{2-\mathrm{x}}^{\mathrm{x}} \\ =\frac{1}{2}{\int }_1^2\mathrm{x}({\mathrm{x}}^2-4+4\mathrm{x}-{\mathrm{x}}^2)\mathrm{dx} \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over x:

$$\begin{gathered} |=2{\overline{)\left(\frac{x^3}{3}-\frac{1}{2}{x}^2\right)}}_1^2 \\ =\frac{5}{3} \end{gathered}$$

Therefore, the value is$\frac{5}{2}$